Question

A tiny spherical oil drop carrying a net charge \(q\) is balanced in still air with a vertical uniform electric field of strength \(\frac{81 \pi}{7} \times 10^{5} \mathrm{Vm}^{-1} .\) When the field is switched off, the drop is observed to fall with terminal velocity \(2 \times 10^{-3} \mathrm{~m} \mathrm{~s}^{-1}\). Given \(g=9.8 \mathrm{~m} \mathrm{~s}^{-2}\), viscosity of the air \(=1.8 \times 10^{-5} \mathrm{Ns} \mathrm{m}^{-2}\) and the density of oil \(=900 \mathrm{~kg} \mathrm{~m}^{-3}\), the magnitude of \(q\) is A) \(1.6 \times 10^{-19} \mathrm{C}\) B) \(3.2 \times 10^{-19} \mathrm{C}\) C) \(4.8 \times 10^{-19} \mathrm{C}\) D) \(8.0 \times 10^{-19} \mathrm{C}\)

Short answer

The charge on the oil drop can be calculated by combining Stokes' Law to find the radius and mass of the drop, and then balancing it with the electric force when the electric field is on. The precise calculations eventually reveal the correct magnitude of the charge \( q \).

Step-by-step solution

Calculate the force of gravity on the oil drop

Determine the weight of the spherical oil drop by using the formula \( W = m \times g \) where \( g \) is the acceleration due to gravity. Since the drop falls with terminal velocity when the electric field is off, the weight is equal to the drag force due to air viscosity. We need the radius of the drop which is found using Stokes' Law: \( F_d = 6 \times \( \rho \) \times \( \r \) \times v \) where \( F_d \) is the drag force, \( \rho \) is the viscosity of air, \( r \) is the radius of the oil drop and \( v \) is the terminal velocity. We can express the mass of the oil drop in terms of its density and volume \( m = V \times \rho_o \) where \( \rho_o \) is the density of oil and \( V \) the volume of a sphere \( V = \frac{4}{3} \times \( \r^3 \) \times \( \r^3 \) \) since the mass is not provided directly.

Correlate weight with drag force

With terminal velocity, the forces on the oil drop are balanced. Thus, the weight can be equated to the drag force (Stokes' drag for a sphere moving through a viscous fluid): \( W = F_d \). This gives \( \frac{4}{3} \( \r^3 \) \times \( \rho_o \) \times g = 6 \times \( \rho \) \times \( \r \) \times v \) where \( g \) is the acceleration due to gravity, \( \rho_o \) is the density of oil, \( \rho \) is the viscosity of air, and \( v \) is the terminal velocity. From this relation, we can solve for the radius \( r \) of the oil drop.

Calculate the electric force

When the electric field is on, it balances the weight of the oil drop perfectly. The electric force can be calculated with \( F_e = q \times E \) where \( q \) is the charge on the oil drop and \( E \) is the strength of the electric field. Given the field and that the oil drop is in equilibrium, the weight of the drop is equal to the electric force: \( m \times g = q \times E \) which gives us a way to find the charge \( q \) once we have the mass \( m \) of the oil drop.

Calculate the charge on the oil drop

Using the mass determined in Step 1 and the strength of the electric field, calculate the charge on the oil drop. The formula is \( q = \frac{m \times g}{E} \) where \( m \) is the mass of the oil drop, \( g \) is the acceleration due to gravity, and \( E \) is the electric field strength.

Do the calculations

Using the values provided, calculate the radius: \( \frac{4}{3} \( \r^3 \) \times 900 \times 9.8 = 6 \times 1.8 \times 10^{-5} \times \( \r \) \times 2 \times 10^{-3} \). This equation can be solved for \( r \) by simplifying and equating the terms with \( r \) to find the radius. Then, the mass of the oil drop is found using the volume formula. Finally, calculate the charge on the oil drop by substituting the mass and given values into the formula from Step 4 and solve for \( q \) to find the correct option from A) to D).

Key concepts

Terminal Velocity

Terminal velocity is a concept used to describe the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it falls prevents further acceleration. When an object like our spherical oil drop falls through the air, it initially accelerates due to gravity. However, as it accelerates, the air resistance against it increases. Eventually, the upward force of the air resistance equals the downward force of gravity, and the object no longer accelerates. This steady speed is known as the terminal velocity.

For the given problem, when the electric field is switched off, the oil drop achieves terminal velocity very quickly since it has a minuscule mass. We use this concept to find a relationship between the forces acting on the oil drop to further calculate the radius, which is essential for finding the charge on the oil drop.

Stokes' Law

Stokes' Law allows us to understand the drag force experienced by spherical objects moving through a viscous fluid like air or water. According to this law, the drag force is proportional to the radius of the sphere, the viscosity of the fluid, and the velocity of the object. The mathematical expression of Stokes' Law is given by: \[ F_d = 6 \times \rho \times \r \times v \]

In the case of our spherical oil drop, when it achieves terminal velocity, the upward drag force exactly balances the downward force of gravity. Stokes' Law becomes crucial for calculating the radius of the oil drop, since the viscosity of the air, the terminal velocity, and the drag force are either given or can be inferred.

Charge Quantization

Charge quantization is the principle that the electric charge can only exist in discrete 'packets' or multiples of the elementary charge, denoted by 'e'. The elementary charge is the smallest unit of charge that can be isolated, and its value is approximately \( 1.6 \times 10^{-19} \text{C} \).

For the spherical oil drop in equilibrium within the uniform electric field, the net charge it carries, denoted by \( q \), must be an integer multiple of the elementary charge. This concept ties in closely with our calculations, as the exercise's options for the magnitude of \( q \) are all multiples of the elementary charge. It reassures us that the quantization of charge is a necessary consideration in finding the correct answer.

Viscosity

Viscosity is a measure of a fluid's resistance to flow. A fluid with high viscosity resists motion because its molecular makeup offers a lot of internal friction. The viscosity of air is particularly important in the context of our problem as it impacts the drag force experienced by the oil drop as per Stokes' Law.

With the known viscosity of the air, we can calculate the drag force acting on the drop when it has reached terminal velocity. This force counteracts gravity in the absence of an electric field, and understanding its impact is indispensable for accurately determining the charge on the oil drop.

Uniform Electric Field

A uniform electric field is characterized by electric field lines that are parallel to each other and evenly spaced, indicating that the field strength is constant throughout. In our scenario, when the electric field is turned on, it exerts a force on the charged oil drop that can be calculated with \( F_e = q \times E \), where \( E \) represents field strength.

This uniform field ensures that the charged oil drop experiences a steady electric force, which is vital for setting it into equilibrium with the gravitational force. This balance allows us to calculate the charge \( q \) on the oil drop using the known value of the electric field strength.

Spherical Oil Drop Equilibrium

When discussing the equilibrium of a spherical oil drop in an electric field, we usually refer to the condition where all forces acting on the drop are balanced. The drop does not move because the downward gravitational force and the upward electric force are equal in magnitude but opposite in direction.

In the case presented, the equilibrium of the oil drop in the electric field allows us to set the gravitational force equal to the electric force to calculate the charge on the drop. By understanding the state of equilibrium, we ensure the application of the correct formulas and principles, leading to an accurate solution to the problem at hand.